Swift 4 Error: "Cannot convert value of type 'URLRequest' to expected argument type 'URL!'"

Question

Just what the title says. I am getting this error thrown and not sure if this has to do with the Swift 4 conversion. I have been trying to work around the code but no luck yet so decided to throw this question up on SO. This error is getting thrown on line: "cell.businessPostImage.setImageWith(postRequest,". Thanks.

cell.businessPostImage.image = nil if let postURL = URL(string: downloadURL) { let postRequest = URLRequest(url: postURL) cell.businessPostImage.setImageWith(postRequest, placeholderImage: nil, options: { (imageRequest, imageResponse, image) in cell.businessPostImage.contentMode = UIViewContentMode.scaleToFill cell.businessPostImage.image = image }, completed: { (imageRequest, imageResponse, error) -> Void in // failure downloading image print("Error downloading Firebase post image") print(error) }) }

Please check the picture I just added of the new error getting thrown because of the edit. — Lukas Bimba
– Lukas Bimba, Commented Oct 4, 2017 at 21:00
You are trying to pass a closure to the options parameter. — rmaddy
– rmaddy, Commented Oct 4, 2017 at 21:04
I understand, should I reformat this block of code? or is there a solution for this? — Lukas Bimba
– Lukas Bimba, Commented Oct 4, 2017 at 21:07

Vini App · Accepted Answer · 2017-10-04 22:59:26Z

You have to use like this :

if let postURL = URL(string: downloadURL) { let postRequest = URLRequest(url: postURL) cell.businessPostImage.setImageWith(postURL, placeholderImage: nil, options: SDWebImageOptions.ProgressiveDownload, completed: { (imageRequest, imageResponse, error) -> Void in // failure downloading image print("Error downloading Firebase post image") print(error) }) }

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Swift 4 Error: "Cannot convert value of type 'URLRequest' to expected argument type 'URL!'"

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